Hydrofluoric acid is a weak monoprotic acid with Ka = 6.6×10-4 M.
1)Calculate the pH of a 6.99×10-2 M solution of hydrofluoric acid.
2)Calculate the pH of a solution that contains 6.99×10-2 M hydrofluoric acid and 5.59×10-2 M sodium fluoride.
HF => H+ + F- ; Ka=6.6*10^-4
(6.6*10^-4) = (x^2) / (0.0699 - x) [rearrange and quadratic formula]
=> x = 6.5 * 10^-3 M = [H+]
so pH = -log(6.5 * 10^-3) = 2.19
pH = pKa + log[base]/[acid]
pH = -log(6.6*10^-4) + log[(0.0559)/(0.0699)] = 3.08
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